Integral log Gamma 0 to 1/2

In this post we will proof the following two integrals related to the derivative of the Riemann Zeta function at 2


\int_0^{1/2}\ln\left(\Gamma(x)\right)dx=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi


\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A

Recall Kummer´s fourier expansion for LogGamma   0<x<1


\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right)(1)

Integrating (1) from 0 to x


\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2}(2)

Letting x=\frac12 in (2)


\begin{aligned} \int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\ &=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\ &=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\ &=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\ &=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare \end{aligned}

Where we used that 

\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)

Proved here, and that

\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)

For the second integral, lets first evalute the following integral


                                              \begin{aligned} I(s)&=\int_0^\infty \frac{x^{s-1}}{e^{cx}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{e^{-x}x^{s-1}}{1-e^{-x}}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \, \int_0^\infty x^{s-1}e^{-kx}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{\Gamma(s)\zeta(s)}{c^s} \end{aligned}

Then, differentiating both sides w.r. to s


\begin{aligned} \int_0^\infty \frac{x^{s-1} \ln x}{e^{cx}-1}\,dx&=\frac{d}{ds}\left(\frac{\Gamma(s)\zeta(s)}{c^s} \right)\\ &=\frac{c^s\left(\Gamma^{\prime}(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-c^s\ln c \left( \Gamma(s)\zeta(s) \right)}{(c^s)^2}\\ &=\frac{\left(\Gamma(s)\psi(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-\ln c \left( \Gamma(s)\zeta(s) \right)}{c^s}\\ \end{aligned}

Setting s=2


\begin{aligned} \int_0^\infty \frac{x \ln x}{e^{cx}-1}\,dx&=\frac{\left(\Gamma(2)\psi(2)\zeta(2)+\Gamma(2)\zeta^{\prime}(2) \right)-\ln c \left( \Gamma(2)\zeta(2) \right)}{c^2}\\ &=\frac{\left((1-\gamma)\zeta(2)+\zeta(2)\left(\ln 2\pi+\gamma-12 \ln A\right) \right)-\zeta(2)\ln c }{c^2}\\ &=\zeta(2)\left(\frac{1-\ln c+\ln 2\pi-12 \ln A }{c^2}\right)\\ \end{aligned}

Setting c=\pi

\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A \qquad \blacksquare


Where we again made use of the fact that


\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)


Comments

Post a Comment

Popular posts from this blog

HARD INTEGRAL - PART II

Image
       Today We will compute the following integral following the same ideas of the previous post : Recall (see here ) (1) Integrating both sides of (1)  from 0 to Computing    Recall (see here ) Letting we obtain We are looking for the Imaginary part of the equation above: Computing the quantities: The Glaisher function We know that (see here ): If we integrate from 0 to x we obtain Integrating from 0 to x we obtain
Read more

Fourier series of Loggamma function - Part 2

Image
                        This post is part 2 of this post , where we started to derive the fourier series of the Loggamma function. Today we are going to evaluate the coefficient   and then, put all the pieces together to find the final expression.  Evaluation of   Consider Integrating by parts now observe the following with regards to the limit: therefore (1) Recall now the integral representation of the Digamma function (2) Plugging (2) on the right hand side of (1) (3) Lets now concentrate in the inner integral in blue and break it down in two integrals, namely Lest begin with (4) Now This integral was already evaluated here , therefore (5) Plugging (4) and (5) back in (3), we get (6) We will consider each of these integrals separately To compute     we perform an integration by parts (7) Where in the last line we used the resul...
Read more

A very useful integral representation of the Digamma function

Image
In this post I will derive an Integral representation for the digamma function which will turn out to be very useful in the computation of the coefficients of the Fourier series of the function that I plan to derive soon in a future post. In these notes I try as much as possible to make each one of them as self content as possible, therefore, although it will be a little long, I will start this derivation almost from scratch, i.e., I will start proving first a product representation for the gamma function . Ok, enough talking, lets do some math! 1. Gauss product for the Gamma function The Gauss product for the Gamma function is given by the following formula (1.1) To prove this formula, recall the well known fact that (1.2) and the standard integral representation of the Gamma function that we are assuming here as the definition of the Gamma function (1.3) substituting (1.2) in (1.3) and taking the limit , we can rewrite as Now lets consider the following integral and...
Read more